Have a proof re. cyclic groups, need a little explaining

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Homework Help Overview

The discussion revolves around a proof related to the properties of elements in a group, specifically focusing on the relationship between the orders of the products of elements in a group. The context is set within the study of cyclic groups.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the order of the product of elements and question whether the group must be cyclic or abelian for the proof to hold. There is an examination of the steps in the proof and the reasoning behind regrouping terms in the absence of commutativity.

Discussion Status

Participants are actively engaging with the proof and clarifying their understanding of the concepts involved. Some guidance has been offered regarding the validity of the proof regardless of whether the group is abelian, and there is a recognition of the need for careful reasoning about the properties of group elements.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the group structure, particularly whether it is cyclic or abelian, and how these assumptions affect the proof. Participants express confusion about the manipulation of terms without commutativity being assumed.

bennyska
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Homework Statement


Let a,b be elements of a group G. show that if ab has finite order, then ba has finite order.


Homework Equations





The Attempt at a Solution


provided proof:
Let n be the order of ab so that (ab)n = e. Multiplying this equation on the left by b and on the right by a, we find that (ba)n+1 = bea = (ba)e. Cancellation of the first factor ba from both sides shows that (ba)n = e, so the order of ba is < n. If the order of ba were less than n, a symmetric argument would show that the order of ab is less than n, contrary to our choice of n. Thus ba has order n also.

okay. so, this is from a chapter on cyclic groups, so I'm assuming that it has to do with, duh, cyclic groups. i know cyclic groups are abelian, and if that were to be assumed from the beginning, i believe this problem would be relatively easy. although, actually, it probably wouldn't even be a problem, since it would be true from the beginning (since we could just say ab=ba, and abn=ban

so:
Let n be the order of ab so that (ab)n = e.
got that, that makes sense, i believe that's the definition of order of an element.
WAIT. here's the crux, something i (maybe) just realized as i typed this sentence. since we define the order of ab such that abn=e, does that imply that G is cyclic, and thus abelian? that would make things easier, if I'm right. so far, a quick internet search has failed me on an answer.
 
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If the group is abelian then ab=ba and there is really nothing to prove. The proof you gave holds regardless of whether the group is abelian or not.
 
okay, i was just confused as how b(ab)na could turn into (ba)n+1 without it being abelian. i see how it's from letting (ab)n = e. thanks.
 
bennyska said:
okay, i was just confused as how b(ab)na could turn into (ba)n+1 without it being abelian. i see how it's from letting (ab)n = e. thanks.

It's actually just from regrouping the terms.
 
yeah, actually, i still don't get it. how can you regroup the terms if it's not commutative? or at least it's not assumed to be commutative.
we have (ab)(ab)...(ab) n times. so b(ab)na = b(ab)(ab)...(ab)a...
Oh! got it! thanks!
 

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